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Let $f\colon R/L \to R$ be such a right inverse, so that $\pi \circ f = \operatorname{id}_{R/L}$. I'll use a bar to denote the image of elements of $R$ in $R/L$. From $(\pi \circ f)(\bar1) = \bar1$ it follows that $f(\bar 1) = 1 + x$ for some $x \in L$. Select a $y \in L - L^2$. What can you say about $f(\bar y) = f(\bar 0) = 0$?